NEET Physics Simple Harmonic Motion MCQ Question with Solution

A block attached to a spring with spring constant 50 N/m oscillates with an amplitude of 0.1 m. What is the total energy of the system at maximum displacement?

A

0.25 J

B

0.125 J

C

0.5 J

D

0.05 J

Correct Answer

Option A

Detailed Explanation

The total energy in simple harmonic motion is given by the potential energy at maximum displacement. Using the formula $\text{PE} = \frac{1}{2}kx^2$ , where $k = 50 \text{ N/m}$ and $x = 0.1 \text{ m}$ , the energy is $0.25 \text{ J}$ .

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